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Margaret [11]
3 years ago
7

A block of ice at 0°C is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°C. If all but 2.0 g of ice me

lt, what was the original mass of the block of ice?
Physics
1 answer:
Over [174]3 years ago
7 0

Answer:

original mass of the block of ice is 38.34 gram

Explanation:

Given data

cup mass = 150 g

ice temperature = 0°C

water mass = 210 g

water temperature = 12°C

ice melt = 2 gram

to find out

solution

we know here

specific heat of aluminum is c = 0.900 joule/gram °C

Specific heat of water C =  4.186 joule/gram °C

so here temperature difference is dt =  12- 0 = 12°C

so here heat lost by water and cup are given by

heat lost  = cup mass × c  × dt + water  mass × C × dt

heat lost  = 150 × 0.900  × 12 + 210 × 4.186 × 12

heat lost  = 12168.72 J

so

mass of ice melt here = heat lost / latent heat of fusion

here we know latent heat of fusion = 334.88 joule/gram

so

mass of ice melt  =  12168.72 / 334.88

mass of ice melt  is 36.337554 gram

so mass of ice is here = mass of ice melt + ice melt

mass of ice  =  36.337554 + 2

mass of ice  =  38.337554 gram

so original mass of the block of ice is 38.34 gram

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musickatia [10]

Answer:

Number of electrons, n=5.62\times 10^{21}

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, I=\dfrac{q}{t}

I\times t=n\times e

n=\dfrac{It}{e}

e is the charge of an electron

n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}

n=5.62\times 10^{21}

So, the number of electrons pass through the resistor is 5.62\times 10^{21}. Hence, this is the required solution.

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3 years ago
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Answer:

C. it will not change.

Explanation:

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Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the
Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) ,   is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

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T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

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A=0.37 m

6 0
3 years ago
Really easy it's physics.
iragen [17]
I think it’s B because we aren’t looking for Vf
4 0
3 years ago
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